# The Snuark Algebra as a QFT

The snuark algebra is simply a finite subset of the Pauli algebra, the subset generated by spin in the +x, +y, and +z directions. If we were to do our calculations by the usual methods of the Pauli algebra, we would choose an orientation, typically +z, and define the +x and +y states in terms of the +z (spin up) and -z (spin down) states.

Our objective in this post is to relate the snuark algebra to the more traditional ways of calculating with the Pauli algebra. We will do this with the objective of turning the snuark algebra into a QFT, just as qubits can be turned into a QFT. And it will be the QFT theory of snuarks that will allow us to derive the Koide mass formulas. Our primary application of the snuark algebra is to the particle masses. Even virtual particles have masses, so as far as a QFT goes, to understand mass we need only describe the QFT for the virtual particles. This simplifies things considerably because virtual particles are naturally described in a density matrix form.

In the qubit QFT, the propagators are density matrix states like (1+x)/2. Two oppositely oriented such states add to unity, for example, (1+x)/2 + (1-x)/2 = 1. We can insert this unity into the middle of a propagator without changing the results of the calculation.

For example, let the propagator be (1+z)/2. Then
(1+z)/2 = (1+z)/2 (1+z)/2
= (1+z)/2 [ (1+x)/2 + (1-x)/2 ] (1+z)/2
= (1+z)/2 (1+x)/2 (1+z)/2 + (1+z)/2 (1-x)/2 (1+z)/2
= 0.5(1+z)/2 + 0.5(1+z)/2.
The last line can be obtained in a number of ways. The one most familiar to the audience is to substitute in the Pauli algebra for the density matrices: Written as Feynman diagrams, putting 1 = (1+x)/2 + (1-x)/2 in the middle of (1+z)/2 amounts to splitting the (1+z)/2 propagator into two halves interupted by the (1+-x)/2 propagators: These examples are unphysical in that we haven’t given any physical reason why the particle should have its polarity changed. In the physical world, when the orientation of a particle is changed like this, a gauge boson needs to be associated, and a vertex needs to be included in the Feynman diagram. In the above, the vertex was chosen to be 1 to allow the equality to work.

To make a physically realistic reason for a propagator to switch polarity, we need to introduce a gauge boson. Following the example of QED, we will use a vertex of amplitude +- iq, where q is a real number that defines the strength of the interaction. The sign of the interaction will depend on the orientations of the initial and final propagator as follows: The signs for the process (1+x)/2 to (1+y)/2 is the negative of the sign for the process (1+y)/2 to (1+x)/2 by Hermiticity. The rest of the signs follow by rotating through the even permutations of x, y, and z and so are required by symmetry. Putting this into Feynman diagram form, along with the algebraic notation I prefer and the Pauli notation that is standard in the industry, we have: These are the basic diagrams that we will use to derive the Koide mass formulas. To reduce products, we will find it very useful to know the following: The above can be shown by replacing the projection operators with Pauli matrices. We will need it, along with the various results you get by permuting x, y, and z around. The even permutations have the same form, while the odd permutations take a minus sign on the imaginary unit.

The above formula is a specific case of a more general result for three spin projection operators (not necessarily perpendicular as in the above case). The product reduces to the first and last operators, times a complex number whose magnitude is given by the familiar sqrt((1 + cos(theta))/2) formula of particle physics, and whose phase is given by half the surface area of the oriented spherical triangle defined by the three spin projection operators.

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### 5 responses to “The Snuark Algebra as a QFT”

1. kneemo

How do the 3×3 lepton mass matrices fit into this framework?

2. carlbrannen

Kneemo,

The next step is to write the propagators for a simple bound state (i.e. a lepton) in terms of the Feynman diagrams for the snuarks of which it is composed. When you organize the Feynman diagrams into groups that feed into each other, it becomes natural to assemble them into a matrix form.

This all works because it is a qubit QFT and works to all orders in perturbation theory. But we will be looking at it from the point of view of making the Feynman diagrams fundamental rather than a Lagrangian. You can recast it the traditional way on your own if you want. Either way, you end up with a set of Feynman diagrams that you assemble to make interactions.

Taking the matrix analogy further requires replacing the Feynman diagrams with complex numbers. To do that you have to add the complex phases that take into account how the snuark propagators interact with each other (i.e. that exp (i pi/4) factor), and you have to add complex phases for the graviton interactions, which eventually becomes that damned number.

I think all that will fit into the next blog post. And after that, I’ve got some cool new surprises in the baryon resonances.

3. Kea

And after that, I’ve got some cool new surprises in the baryon resonances.

Oh, goody! I can’t wait! Thanks for the clear exposition. Could we use honeycombs of Feynman diagrams?

4. carlbrannen

Kea, the Feynman diagrams are going to be more linear than 120 degree angled as one would need for a honeycomb.

However, I have a blog post that got left on the cutting room floor that got into honeycombs. Suppose you have a thingy that can travel in the +-x, +-y, and +-z directions. At the outset it is at the origin going in the +x direction. Make the steps each the same length.

At each step, it changes direction. The first rule is that it must always change from a + to a – or back. The second rule is that it cannot completely reverse its direction. So if it starts out at +x, the next step is -y or -z.

If you draw the orbits it can make to get back to the origin going in the +x direction, it will have followed six of the edges of a cube. Projected onto the (1,1,1) direction, the result is a hexagon.

Beginning in the +x direction, there are two possible next steps, -y and -z. They lead to two hexagons that are adjacent. That’s a spin in terms of going around in a circle.

Adjacent hexagons? That’s the start of a honecomb.